A stone of mass 1 kg is thrown with a velocity of `20ms^(-1)` across the frozen surface of a lake and it comes to rest after travelling a distance of 50m. What is the magnitude of the force opposing the motion of the stone?

Options

A. `-2N`
B. `-1N`
C. `-3N`
D. `-4N`

Answer

D

Explanation

Given u=`20ms^(-1)` , v=0 , s=50 m and m=1kg
To calculate force, we have the formula F=ma, but we have to first calculate acceleration a. Using the third equation of motion, i.e.,
`v^(2)=u^(2)+2as`
` (0)^(2)=(20)^(2)+2xxaxx50`
`100a=-400`
`a=-(400)/(100)=-4ms^(-2)`
Acceleration a `=-4ms^(-2)` (-ve sign shows that speed of the stone decrease,i.e., retardation)
Now, `F=ma=1kgxx(-4)ms^(-2)=-4kg ms^(-2)= -4N`
Thus , force of friction between the stone and the ice is -4N.
The negative value of force shows that the frictional force acts in a direction opposite to the direction of motion.